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\lhead{MAT\,137\,Y1Y}
\chead{Problem Set \#\,2}
\rhead{Summer 2014}


\begin{document}

\begin{large}
  \noindent
  Name : Robert Staskiewicz \hfill Student Number: 1000340570\\
  \indent  \hfill Tutorial: 0201
\end{large}

\medskip

\noindent
\rule{\textwidth}{.5pt}

\subsection*{Topic: Limits}

\medskip

\begin{enumerate}
 
 \item
 Evaluate the following limits:     
    % place solution to question 1 below
	\begin{enumerate}
    \item
    \medskip
    \noindent
    Evaluate:
    $$     \lim_{x \to 2} \frac{x^2-4}{x^2-5x+6} $$
    {\bf Proof:} 
    \begin{pindent}
        \begin{assumption}{$ \lim_{x \to 2} \frac{x^2-4}{x^2-5x+6} $}
		Then $ \lim_{x \to 2} \frac{x^2-4}{x^2-5x+6} = \lim_{x \to 2} \frac{(x+2)\cancel{(x-2)}}{(x-3)\cancel{(x-2)}}= \lim_{x \to 2} \frac{x+2}{x-3} $\\
		Then $\lim_{x \to 2} \frac{x+2}{x-3} = \frac{2+2}{2-3}=-4 $\just{$\text{If P(x) is a polynomial, then}\lim_{x \to a} P(x) = P(a)$}
		\end{assumption}
		Therefore, $\lim_{x \to 2} \frac{x^2-4}{x^2-5x+6}=-4$\\
		    \end{pindent}
	\item
		Evaluate: $$\lim_{x \to 2} \frac{x^2-5x+6}{x^2-4x+4}$$
		{\bf Proof:}
		\begin{assumption}{$\lim_{x \to 2} \frac{x^2-5x+6}{x^2-4x+4}$}
				Then $\lim_{x \to 2} \frac{x^2-5x+6}{x^2-4x+4}= \lim_{x \to 2} \frac{(x-3)\cancel{(x-2)}}{(x-2)\cancel{(x-2)}}= \lim_{x \to 2} \frac{(x-3)}{(x-2)} $ \\
				Then $ \lim_{x \to 2} (x-3) = -1 $ \\
				And $ \lim_{x \to 2} (x-2) = 0 $\\
				Then $\lim_{x \to 2} \frac{(x-3)}{(x-2)}\ \text{does not exist.}$ \\ \just{
				$\text{If}\ \lim_{x \to c} = L \ \text{with} \ L \not= 0 \ \text{and} \ \lim_{x \to c} g(x) = 0, \text{then} \lim_{x \to c} \frac{f(x)}{g(x)}\ \text{does not exist.}  $ }
				\end{assumption} 
		Therefore $\lim_{x \to 2} \frac{x^2-5x+6}{x^2-4x+4}\ \text{does not exist.}$

    \medskip

	\item 
		Evaluate: $$\lim_{x \to 1} \frac{x-1}{|x^3-x^2+4x-4|}$$
		{\bf Proof:}
		\begin{assumption}{
		 $\lim_{x \to 1} \frac{x-1}{|x^3-x^2+4x-4|} $}
		 Then $ \lim_{x \to 1} \frac{x-1}{|x^3-x^2+4x-4|} = \lim_{x \to 1} \frac{x-1}{|(x-1)(x^2+4)|}  $
		  \begin{displaymath}
		     |(x-1)(x^2+4)| = \left\{
		       \begin{array}{lr}
		         (x-1)(x^2+4) & : x \geq 1\\
		         -(x-1)(x^2+4) & : x < 1
		       \end{array}
		     \right.
		  \end{displaymath}
		  If $x \geq 1$, then $\lim_{x \to 1^+} \frac{x-1}{(x-1)(x^2+4)}=\lim_{x \to 1^+} \frac{\cancel{(x-1)}}{\cancel{(x-1)}(x^2+4)}= \lim_{x \to 1^+} \frac{1}{(x^2+4)}=\frac{1}{5}$\\
		  If $x < 1 $, then $ \lim_{x \to 1^-} \frac{x-1}{-(x-1)(x^2+4)}=\lim_{x \to 1^-} \frac{\cancel{(x-1)}}{-\cancel{(x-1)}(x^2+4)}= \lim_{x \to 1^-} \frac{1}{-(x^2+4)}=-\frac{1}{5}  $\\
		  Then, the one-sided limits are different.
		\end{assumption}
		Therefore, $ \lim_{x \to 1} \frac{x-1}{|x^3-x^2+4x-4|}\ \text{does not exist.} $
	\newpage
	\item
	Evaluate:
	    $$     \lim_{x \to \infty} \frac{x^3-8x+2}{x^4-5x+6} $$
	    {\bf Proof:} 
	    \begin{pindent}
	        \begin{assumption}{$ \lim_{x \to \infty} \frac{x^3-8x+2}{x^4-5x+6} $}
			Then $ \lim_{x \to \infty} \frac{x^3-8x+2}{x^4-5x+6} = \lim_{x \to \infty} \frac{x^4(\frac{1}{x}-\frac{8}{x^3}-\frac{2}{x^4})}{x^4(1-\frac{5}{x^3}+\frac{6}{x^4}) }= \lim_{x \to \infty} \frac{\cancel{x^4}(\frac{1}{x}-\frac{8}{x^3}-\frac{2}{x^4})}{\cancel{x^4}(1-\frac{5}{x^3}+\frac{6}{x^4}) }$\\
			Then $\lim_{x \to \infty} \frac{\frac{1}{x}-\frac{8}{x^3}-\frac{2}{x^4}}{1-\frac{5}{x^3}+\frac{6}{x^4} } = \frac{0-0-0}{1-0-0}=\frac{0}{1}=0 $ \\
			
			\end{assumption}
			Therefore, $\lim_{x \to \infty} \frac{x^3-8x+2}{x^4-5x+6}=0$\\
			    \end{pindent}
			    
	\item
	Evaluate:
		    $$     \lim_{x \to -\infty} \frac{x^5-6x}{3x^5+5x} $$
		    {\bf Proof:} 
		    \begin{pindent}
		        \begin{assumption}{$\lim_{x \to -\infty} \frac{x^5-6x}{3x^5+5x}$}
				Then $ \lim_{x \to -\infty} \frac{x^5-6x}{3x^5+5x} = 
				\lim_{x \to -\infty} \frac{x^5(1-\frac{6}{x^4})}{x^5(3+\frac{5}{x^4})}  = \lim_{x \to -\infty} \frac{\cancel{x^5}(1-\frac{6}{x^4})}{\cancel{x^5}(3+\frac{5}{x^4})} $ \\
				Then $ \lim_{x \to -\infty} \frac{1-\frac{6}{x^4}}{3+\frac{5}{x^4}}= \frac{1-0}{3+0}=\frac{1}{3} $
				
				\end{assumption}
				Therefore, $\lim_{x \to -\infty} \frac{x^5-6x}{3x^5+5x}=\frac{1}{3}$\\
				    \end{pindent}
    \end{enumerate}

    \item
        	Find the following limits for the given function:
    \begin{enumerate}
        \item
        $ \lim_{x \to -1^+} f(x) = -\infty $\just{The function decreases from the right without bound.} \\
        \item
        $ \lim_{x \to -1^-} f(x) = \infty $ \just{The function increases from the left without at bound.}
        \item
        $  \lim_{x \to -1} f(x) = \text{does not exist.} $\just{Left limit $\not=$ Right limit; by (a) and (b)}
        \item
        $  \lim_{x \to 1} f(x) = \text{does not exist.} $ \just{$\lim_{x \to 1^-} f(x)=-1$ and $\lim_{x \to 1^+} f(x)=0 $} 
        \item
        $ \lim_{x \to 1.5^+} f(x)=-1 $
        \item
        $  \lim_{x \to 2.5} f(x)=1 $ \just{$\lim_{x \to 2.5^-} f(x)=1$ and $\lim_{x \to 2.5^+} f(x)=1 $}
        \item
        $ \lim_{x \to 1.5} f(f(x))= f(\lim_{x \to 1.5}f(x)) $\just{We use without proof: $\lim_{x \to c}f(g(x))=f(\lim_{x \to c}g(x))$}\\
        Then $ \lim_{x \to 1.5}f(x) = -1 $ \just{but $f(x)$ is always greater than $-1$}\\
        Then we have $ f(\lim_{x \to 1.5}f(x)) = -\infty $\\
        Therefore, $ \lim_{x \to 1.5} f(f(x)) = -\infty $
        \item
        $ \lim_{x \to 0} f(1+x) = 0 $\just{We can think of this as $f(1+\epsilon)$}
        \item
        $ \lim_{x \to 0} f(1+x^2) = 0 $\just{Similar reasoning to (h) }
        
        
        
        \medskip
        \noindent
    \end{enumerate}

%    
%    % place solution to question 2 below
%
%       
%    \item
%    \begin{enumerate}
%    \item
%    Consider the following claim:
%       \begin{quote} 
%       "A function $f$ is periodic if and only if there exists a $k > 0$ such that for every $x,f(k + x) = f(x)$".\\
%       \end{quote}
%       P is a predicate that operates on the set $\R$ and is defined by P(x): Function x is periodic.\\
%       We can express this in the notation of symbolic logic as:
%       $$    P(x) \iff \forall x \in \R, \exists k \in \R, k>0,  f(k+x) = f(x)  $$
%       The negation of a biconditional is: $\neg(P \iff Q)\ \iff [(P \land \neg Q) \lor (Q \land \neg P)]:$\newline \small $[\neg P(x) \land[ \forall x \in \R, \exists k \in \R, k>0,  f(k+x) = f(x)  ]] \lor [P(x) \land [\exists x \in \R, \forall k \in \R, k\leq 0, f(k+x)\not=f(x) ]]$
%       
%       \item
%       
%       \end{enumerate}
%       
%           Consider the following claim:
%           \begin{quote}
%           Prove the following by contradiction: If $||x|-|y||<|x-y|, \text{then}\  xy < 0$.
%           \end{quote}
%       {\bf Proof:} \\
%    We can express this in the notation of symbolic logic as:
%    $$\forall x,y \in \R,\ (||x|-|y||<|x-y|) \implies (xy<0) $$ \\
%		{\bf Proof:} \\
%		    \begin{pindent}
%		
%          \begin{assumption}{$||x|-|y||<|x-y|$ is true}
%				Suppose $xy \geq 0$\\
%				Then $(x > 0 \land y >0) \lor (x< 0 \land y < 0)$\\\\
%				Case1:\\
%				If $(x > 0 \land y >0)$\\
%				Then $(||x|-|y||<|x-y|) \implies(|x-y|<|x-y|)$\just{We reach a contradiction.}\\\\
%				Case2:\\
%				If $(x< 0 \land y < 0)$\\
%				Then $(||x|-|y||<|x-y|) \implies (|-x+y|<|x-y|)$\\
%				\hspace*{10mm}Subcase1:\\
%				\hspace*{10mm}If $x>y$\\
%				\small\hspace*{10mm}Then $(|-x+y|<|x-y|) \implies (-(-x+y)<(x-y))$\just{$-x+y$ is negative because $x>y$}  \\
%				\hspace*{10mm}Then $(|-x+y|<|x-y|) \implies ((x-y)<(x-y))$\just{We reach a contradiction.}\\
%				\hspace*{10mm}Subcase2:\\
%				\hspace*{10mm}If $y>x$\\
%				\hspace*{10mm}Then  $(|-x+y|<|x-y|) \implies ((-x+y)<(-x+y))$\just{We reach a contradiction.}\\
%				Therefore, if $||x|-|y||<|x-y|$ is true, then $xy \geq 0 $ cannot be true.\\
%          \end{assumption}
%          Therefore, $\forall x,y \in \R,\ (||x|-|y||<|x-y|) \implies (xy<0) $
%           
%      \end{pindent}
%      \medskip
%
%	\item
%	    \begin{quote}
%	     Solve for $x: x^2 - 9x + 20 < 0$
%	    \end{quote}
%			
%			 \begin{assumption}{$x \in \R $ and $x^2-9x+20<0$ }
%			 Then $x^2-9x+20=(x-4)(x-5)$ \just{Factoring.}\\
%			 Then $(x-4)(x-5)<0 $\just{Roots at $4 \ \text{and} \ 5$.}\\
%			 We test the intervals $(-\infty,4),\ (4, 5) ,\ (5, \infty)$\\
%			 let $x=0$, then $(0^2-9*0+20=20)$\just{We exclude $(-\infty, 4)$}\\
%			 let $x=4.5$, then $(4.5^2)-9(4.5)+20=-0.25$\just{We include $(4,5)$}\\
%			 let $x=6$, then $6^2-9(6)+20=2$\just{We exclude $(5, \infty)$}\\
%			 Therefore, $\forall x \in \R, ((x^2 - 9x + 20) < 0) \implies \{x\in \R| 4<x<5\} $
%			 \end{assumption}
%			 \newpage
%	\item
%	\begin{enumerate}
%		\item
%		    \begin{quote}
%		     Find the lub and glb (if they exist) for $S = \{x \in R : |4x + 1| < 1\}$\\
%		    \end{quote}
%				First we solve the inequality:
%				 \begin{assumption}{$x\in \R$ and $|4x+1|<1$}
%				 If $|4x+1|>0$\\
%				 Then $(4x+1)<1$\just{Definition of Absolute value.} \\
%				 Then $x<0$\just{Simplifying}\\
%				 If $|4x+1|<0$\\
%				 Then $(-4x-1)<1$\just{Definition of Absolute value.} \\
%				 Then $x>- \frac{1}{2}$\just{Simplifying.}\\
%				 Combining Inequalities, we have $- \frac{1}{2} < x < 0$\\
%				 Therefore, we have a least-upper bound of 0 and a greatest-lower bound of $- \frac{1}{2}$
%				 \end{assumption}
%				 
%		\item
%		
%		\begin{quote}
%				     Prove that if S has a greatest element b, then b = lub S.
%				    \end{quote}
%						 \begin{assumption}{$S$ has a greatest element b and $ b \not= \text{lub} S$}
%						 Suppose $\exists c \in S, (c > b) \implies (c = \text{lub} S) $\small \just{if $b \not=$ lub $S$, then c must be strictly greater than b.}\\
%						 But, we know that b is the greatest element in b.\\
%						 This contradicts our assumption that b is not a lub for $S$.
%
%						 \end{assumption}
%						 Therefore, if S has a greatest element b, then b = lub S.
%		\end{enumerate}
%	\item
%	\begin{enumerate}
%	\item Suppose $a,b,c,d$ are such that $ad-bc\not=0$. Let $f(x)= \frac{ax+b}{cx+d}$. Find $f^{-1}$.\\
%	Let $f(x)=y$\\
%	Then $y=\frac{ax+b}{cx+d}$\\
%	Then $x=\frac{ay+b}{cy+d}$\just{Solving for inverse.}\\
%	Then $x(cy+d)=ay+b	$\\
%	Then $xcy+dx=ay+b $\\
%	Then $xcy-ay=b-dx $\\
%	Then $y(xc-a)=b-dx $\\
%	Then $y=\frac{b-dx}{xc-a} $\\
%	Therefore, $f(x)^{-1}=\frac{b-dx}{xc-a} $
%	
%	\item
%	Let $f(x) = 3x + 4$ and $g(x) = (x + 2)^{\frac{1}{4}}$. Find the inverses of $f, g$ and $g \circ f$ on their domain. Compose $f^{-1} \circ g^{-1}$ and check that $(g \circ f)^{-1} = f^{-1} \circ g^{-1}.$\\\\
%	Assume $f(x)=3x+4 $\\
%	Let $f(x)=y$\\
%	Then $y=3x+4 $\\
%	Then $x=3y+4 $\\
%	Then $x-4=3y $\\
%	Then $\frac{x-4}{3}=y $\\
%	Then $f(x)^{-1}= \frac{x-4}{3}$\\
%	\\
%	Assume $g(x)=(x+2)^{\frac{1}{4}} $\just{Note Domain: $x \geq -2 $ and Range: $ y \geq 0$}\\
%	Let $g(x)=y $\\
%	Then $y=(x+2)^{\frac{1}{4}} $\\
%	Then $x=(y+2)^{\frac{1}{4}} $ \\
%	Then $x^{4}=y+2 $\\
%	Then $x^{4}-2=y $\\
%	Then $g(x)^{-1}=x^{4}-2$\just{Inverse is only defined for $x\geq 0$}\\\\
%	
%	Assume $f(x)=3x+4$ and $g(x)=(x+2)^{\frac{1}{4}} $\\
%	Then $g \circ f  =((3x+4)+2)^{\frac{1}{4}} $ \\
%	Then $g \circ f=(3x+6)^{\frac{1}{4}}  $\just{Note Domain: $x \geq -2$ and Range: $y \geq 0$}\\
%	Let $g \circ f= y$\\
%	Then $y=(3x+6)^{\frac{1}{4}} $\\
%	Then $x^4=3y+6 $ \\
%	Then $x^4-6=3y $ \\
%	Then $\frac{x^4-6}{3}=y $\\
%	Then $(g \circ f)^{-1}= \frac{x^4-6}{3} $\just{Inverse only defined for $x \geq 0$}\\\\
%	Assume $f^{-1}= \frac{x-4}{3} $ and $g^{-1}= x^{4}-2$\just{Defined above.}\\
%	Then $f^{-1} \circ g^{-1}= \frac{(x^4-2)-4}{3}$\\
%	Then $f^{-1} \circ g^{-1}= \frac{x^4-6}{3} $\\
%	And $(g \circ f)^{-1}= \frac{x^4-6}{3} $ \just{Defined above.}\\
%	Then $(g \circ f)^{-1}=f^{-1} \circ g^{-1}  $
%	\end{enumerate}
%	\item
%		\begin{enumerate}
%		\item 
%		Find the domain and the range of $f(x) =\frac{\sqrt{x^2-1}}{x-|x|} $\\
%		Assume $x \geq 0$\\
%		Then $f(x)=\frac{\sqrt{x^2-1}}{x-x} $\\
%		Then $f(x)=\frac{\sqrt{x^2-1}}{0}  $\\
%		Then $f(x)$ is never defined.\\
%		Assume $x < 0$\\
%		Then $f(x)=\frac{\sqrt{x^2-1}}{x+x}  $\\
%		Then $f(x)=\frac{\sqrt{x^2-1}}{2x} $ \just{We have a domain of $(-\infty,0)$}\\
%		However, we must consider the numerator term $\sqrt{x^2-1} $\\
%		Which is defined when $x^2-1\geq 0$ or when $(x \geq 1) \lor (x \leq -1) $\\
%		Then we have $(-\infty,0) \cap (-\infty, -1]$, which is just $(-\infty,-1]$\\
%		For the Range, we can find Domain of the inverse of $f(x)=\frac{\sqrt{x^2-1}}{2x}$\\
%		Then $x=\frac{\sqrt{y^2-1}}{2y} $ \\
%		Then $4x^2y^2 = y^2-1 $\\
%		Then $1=y^2-4x^2y^2,$
%		Then $y^2(1-4x^2)=1,\text{Then } y^2= \frac{1}{1-4x^2}$\\ 
%		Then $y=\pm \frac{1}{\sqrt{1-4x^2}} $\\
%		Therefore, the Domain of the inverse is only defined when $(1-4x^2)\geq 0$, so $ (\frac{1}{2} \geq x) \lor (x \geq -\frac{1}{2})$\\
%		Which gives the inverse function a Domain of $(-\frac{1}{2}, \frac{1}{2}) $\\
%		Therefore, the Range of the original function, $f(x) $ is $(-\frac{1}{2}, \frac{1}{2}) $.\\
%		Therefore, we have a Domain of $(-\infty,-1]$ and a Range of $(-\frac{1}{2}, \frac{1}{2}) $ for $f(x)$
%		\newpage
%		\item
%		 Consider the following function:\
%		  Domain $=  (-\infty, 1)\cup (1, 3) \cup (3, \infty) $ and a Range $= (5, \infty) $\\\\
%		  A function with this Domain and Range is: $ f(x) = (\frac{1}{(x-3)^2}+5)\frac{x^2-4x+3}{(x-1)(x-3)} $\\
%
%		  		    \begin{pindent}
%		  The function appears to be defined at all points but $1$ and $3$.\\
%		  We take a function that has an undefined denominator at $x=1$ and $x=3$,
%		  so we have $(x-3)$ and $(x-1)$ for a denominator $\frac{1}{(x-1)(x-3)}$. However, we want these points to be undefined without asymptotes, so we make the numerator $(x-3)(x-1)$ to get $\frac{(x-3)(x-1)}{(x-3)(x-1)} $.
%		  Then we need a function which has a range strictly greater than 5, we can take $\frac{1}{(x-3)^2}+ 5 $.
%		  Combining all these ideas gives us $ f(x) = (\frac{1}{(x-3)^2}+5)\frac{x^2-4x+3}{(x-1)(x-3)} $
%		\end{pindent}
%		      \medskip
%		\end{enumerate}
\end{enumerate}

\noindent
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